How do you find the asymptotes for #y = x/(x-6)#?

Question

Elias Ackerman · Accepted Answer

The asymptotes are #y=1# and #x=6#

To find the vertical asymptote, we only need to take note the value approached by x when y is made to increase positively or negatively

as y is made to approach #+oo# , the value of (x-6) approaches zero and that is when x approaches +6.

Therefore, #x=6# is a vertical asymptote.

Similarly, To find the horizontal asymptote, we only need to take note the value approached by y when x is made to increase positively or negatively

as x is made to approach #+oo# , the value of y approaches 1.

#lim_(x " "approach +-oo) y=lim_(x " "approach +-oo)(1/(1-6/x))=1#

Therefore, #y=1# is a horizontal asymptote.

kindly see the graph of

#y=x/(x-6)#. graph{y=x/(x-6)[-20,20,-10,10]}

and the graph of the asymptotes #x=6# and #y=1# below. graph{(y-10000000x+6*10000000)(y-1)=0[-20,20,-10,10]} have a nice day!

Elias Ackerman · Answer

undefined To find the asymptotes for $y = \frac{x}{x-6}$, we need to consider both vertical and horizontal asymptotes.

1. **Vertical Asymptotes**: Vertical asymptotes occur where the denominator of the function becomes zero, but the numerator doesn't. So, set the denominator equal to zero and solve for $x$.

$$x - 6 = 0$$
$$x = 6$$

Therefore, there is a vertical asymptote at $x = 6$.

2. **Horizontal Asymptotes**: Horizontal asymptotes occur when $x$ approaches positive or negative infinity. To find horizontal asymptotes, compare the degrees of the numerator and denominator.

In this case, both the numerator and denominator have the same degree (degree 1), so we compare the leading coefficients. The leading coefficient of the numerator is 1, and the leading coefficient of the denominator is also 1.

Therefore, the horizontal asymptote is given by the ratio of the leading coefficients, which is $y = \frac{1}{1} = 1$.

Thus, the asymptotes for $y = \frac{x}{x-6}$ are:
- Vertical asymptote: $x = 6$
- Horizontal asymptote: $y = 1$